Leetcode 15. 3Sum
2018-01-19 13:37
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15. 3Sum
Given an array S of n integers, are there elements a,b, c in S such that a + b + c = 0? Find all unique triplets in the array which gives the sum of zero.
Note: The solution set must not contain duplicate triplets.
For example, given array S = [-1, 0, 1, 2, -1, -4], A solution set is: [ [-1, 0, 1], [-1, -1, 2] ]
题目链接:https://leetcode.com/problems/3sum/description/
题目大意:给定数组S,求和为 0 的三个数的组合,要求组合不重复
解题思路:三重循环时间复杂度是O(n^3),先快速排序,时数组递增有序。
第一层:第一个数按顺序遍历O(n) ;
第二层:第二个数从前往中间遍历,如果总数<0,下标增加,第三个数从最后往中间遍历,如果总数<0,下标增加,O(n) ; 合起来时间复杂度为O(n^2)
代码:
class Solution { public: vector<vector<int> > threeSum(vector<int>& nums) { vector<vector<int> >ans; int n = nums.size(); sort(nums.begin(),nums.end()); for(int i = 0;i < n - 2;i ++){ if(nums[i] > 0){ while(i < n - 3 && nums[i + 1] == nums[i]) i ++; break; } int j = i + 1,k = n - 1; while(j < k){ if(nums[i] + nums[j] > 0) break; while(j < k && nums[i] + nums[j] + nums[k] < 0) j ++; while(j < k && nums[i] + nums[j] + nums[k] > 0) k --; if(j >= k) break; if(nums[i] + nums[j] + nums[k] == 0){ vector <int> cur; cur.push_back(nums[i]); cur.push_back(nums[j]); cur.push_back(nums[k]); ans.push_back(cur); }else{ continue; } while(j + 1 < k && nums[j + 1] == nums[j]) j ++; while(k - 1 > j && nums[k - 1] == nums[k]) k --; j ++; k --; } while(i < n - 3 && nums[i + 1] == nums[i]) i ++; } return ans; } };
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